I have a column with data zeros after 6th column
i want to remove the leading zero after the 6th pipe in the data.
Please let me know if there is any way to do it. I tried to use substr with Trim but its not working.
Let's say your column is called COL something like the following should work:
CONCAT(SUBSTR(1,INSTR(COL,'|', 1,5)), LTRIM(SUBSTR(INSTR(COL,'|', 1,5)+1)),'0'))
Assuming from your current data that you need to only replace every occurrence of |00 with |, You can achieve thing using REPLACE function.
SELECT 'TMB|MLE020828585|74384911WA3S|="''07300058"|74384911|0013' AS Current_String,
replace('TMB|MLE020828585|74384911WA3S|="''07300058"|74384911|0013', '|00', '|') AS result_String
You can replace hard-coded value in above with your column name.
The above query generate result as below.
Current_String | result_String
------------------------------------------------------------------------------------------------------------------------
TMB|MLE020828585|74384911WA3S|="'07300058"|74384911|0013 | TMB|MLE020828585|74384911WA3S|="'07300058"|74384911|13
Hope this will help.
Oleg was correct in utilizing INSTR(), that is how I would do it. He was missing some arguments, though. Also, I wasn't able to get ltrim to work, so I cast it to an int instead. I tested and updated my table with your data using:
UPDATE mytable
SET mycolumn = Substr(mycolumn , 1, Instr(mycolumn , '|', 1, 5))
|| CAST(Substr(mycolumn , Instr(mycolumn , '|', 1, 5) + 1) AS INT)
Related
I have column with software versions. I was trying to remove dot from 2nd occurrence in column, like
select REGEXP_REPLACE('12.5.7.8', '.','');
expected out is 12.578
sample data is here
Is it possible to remove dot from 2nd occurrence
One option is to break this into two pieces:
Get the first number.
Get the rest of the numbers as an array.
Then convert the array to a string with no separator and combine with the first:
select (split_part('12.5.7.8', '.', 1) || '.' ||
array_to_string((REGEXP_SPLIT_TO_ARRAY('12.5.7.8', '[.]'))[2:], '')
)
Another option is to replace the first '.' with something else, then get rid of the '.'s and replace the something else with a '|':
select translate(regexp_replace(version, '^([^.]+)[.](.*)$', '\1|\2'), '|.', '.')
from software_version;
Here is a db<>fiddle with the three versions, including the version a_horse_with_no_name mentions in the comment.
I'd just take the left and right:
concat(
left(str, position('.' in str)),
replace(right(str, -position('.' in str)), '.', '')
)
For a str of 12.34.56.78, the left gives 12. and the right using a negative position gives 34.56.78 upon which the replace then removes the dots
I have column with data like:
'2020193'
'3208391'
'1038291'
'9349203'
The data type is varchar and I can't change it to int (data managed in this datatype always).
I have some rows with trailing spaces like:
' 2222928'
' 3213331'
I need to remove that trailing space from start. I have tried SUBSTRING() or TRIM()/RTRIM()/LTRIM(), but didn't worked any of those.
select (rtrim(ltrim(doc_id))) from bpm.sales where len(doc_id) = 8
select left(doc_id,2) from bpm.sales where len(doc_id) = 8
select charindex(' ',doc_id) from bpm.sales where len(doc_id) = 8
Also, when I am trying to search the data like:
select doc_id from bpm.sales where doc_id = ' 2269203'
I am geting nothing where it exist in the column. With CHARINDEX() I got 0.
Can someone explain me this behaviour and suggest a solution?
You can get rid of everything up to the first character you do want:
select stuff(doc_id, 1, patindex('%[^0-9a-zA-Z]%', doc_id) - 1, '')
i would make a script that iterate over the records of a table with a cursor
and extract from a column value formatted like that "yyy://xx/bb/147011"
only the final number 147011and to put this value in a variable.
It's possible to do something like that?
Many thanks.
You don't need a cursor for this. You can just use a query. The following gets everything after the last /:
select right(str, charindex('/', reverse(str)) - 1 )
from (values ('yyy://xx/bb/147011')) v(str)
It does not specifically check if it is a number, but that can be added as well.
You can also use the below query.
SELECT RIGHT(RTRIM('yyy://xx/bb/147011'),
CHARINDEX('/', REVERSE('/' + RTRIM('yyy://xx/bb/147011'))) - 1) AS LastWord
If numeric value has exact position defined with sample data, then you can do :
SELECT t.*, SUBSTRING(t.col, PATINDEX('%[0-9]%', t.col), LEN(t.col))
FROM table t;
I have a column in table with values similar to this:
key_value
eg:
3933984948498934_khkhk
81299191ahgtyu092092092019_92982
abh182772hjjlj98879bjj_122778999
_ is common in all the values. I need a script to copy some portion of the value i.e. copy everything before _ and not display anything after _ for all the values in that column.
I need a select statement to display output as mentioned above.
eg: Actual value is 3933984948498934_khkhk but I need just 3933984948498934
Actual value 81299191ahgtyu092092092019_92982 but desired output is 81299191ahgtyu092092092019.
I feel using substr function is cumbersome as the values are dynamic.
You could use a regular expression, e.g.:
SELECT REGEXP_SUBSTR(key_value, '^[^_]*')
FROM mytable;
but regular expressions are resource-intensive; you'd be better served using SUBSTR() and INSTR():
SELECT SUBSTR(key_value, 1, INSTR(key_value, '_') - 1)
FROM mytable;
Note that this latter method will fail (actually, it will return NULL) if key_value does not contain an underscore _. So you might wrap it in COALESCE():
SELECT COALESCE(SUBSTR(key_value, 1, INSTR(key_value, '_') - 1), key_value)
FROM mytable;
Using T-SQL, how would I go about getting the last 3 characters of a varchar column?
So the column text is IDS_ENUM_Change_262147_190 and I need 190
SELECT RIGHT(column, 3)
That's all you need.
You can also do LEFT() in the same way.
Bear in mind if you are using this in a WHERE clause that the RIGHT() can't use any indexes.
You can use either way:
SELECT RIGHT(RTRIM(columnName), 3)
OR
SELECT SUBSTRING(columnName, LEN(columnName)-2, 3)
Because more ways to think about it are always good:
select reverse(substring(reverse(columnName), 1, 3))
declare #newdata varchar(30)
set #newdata='IDS_ENUM_Change_262147_190'
select REVERSE(substring(reverse(#newdata),0,charindex('_',reverse(#newdata))))
=== Explanation ===
I found it easier to read written like this:
SELECT
REVERSE( --4.
SUBSTRING( -- 3.
REVERSE(<field_name>),
0,
CHARINDEX( -- 2.
'<your char of choice>',
REVERSE(<field_name>) -- 1.
)
)
)
FROM
<table_name>
Reverse the text
Look for the first occurrence of a specif char (i.e. first occurrence FROM END of text). Gets the index of this char
Looks at the reversed text again. searches from index 0 to index of your char. This gives the string you are looking for, but in reverse
Reversed the reversed string to give you your desired substring
if you want to specifically find strings which ends with desired characters then this would help you...
select * from tablename where col_name like '%190'